If k^3 is divisible by 240 what is the least possible value

Question

If k^3 is divisible by 240 what is the least possible value of integer K?

A. 12
B. 30
C. 60
D. 90
E. 120

Bunuel · Accepted Answer

If k^3 is divisible by 240 what is the least possible value of integer K?

A. 12
B. 30
C. 60
D. 90
E. 120

Make prime factorization:  240=2^4*3*5 .

Now, the powers of each prime of  k^3  must be a multiples of 3 (because of 3), so the least value of  k^3  so that it's divisible by  2^4*3*5  is  k^3=2^6*3^3*5^3  -->  k=2^2*3*5=60 .

Answer: C.

Hope it's clear.

ENGRTOMBA2018 · Answer

Follow posting guidelines (link in my signatures) especially searching for a question before posting a new thread. Additionally, type in the complete question with all the options and do not use pictures to stand for the questions.

Refer above for the solution. For this question, the most straightforward way is to use the options to your advantage.

You know that k^3 must be divisible by 240 ---> You get an INTEGER value when you calculate  k^3/240  ---> this eliminates A) directly as 12 is not divisible by 5 (=a factor of 240).

For B ---> if k =30=2*3*5 --->  k^3 = 2^3*3^3*5^3  and  240 = 2^4*3*5 , you can clearly see that 1 '2' from 240 will not be cancelled out when you write  k^3/240  giving you a NON INTEGER VALUE. Eliminate this option.

For C ---> if  k =60=2^2*3*5  --->  k^3 = 2^6*3^3*5^3  and  240 = 2^4*3*5 , you can clearly see that all the factors of 240 will now be cancelled out when you write  k^3/240  giving you an INTEGER value. This is the correct answer.

As the question was asking about the LEAST value, no need to check D and E.

Hope this helps.

chetan2u · Answer

Hi,
 k^3  is div by 240..
since k is to the power of 3 we will have to prime factorize 240 in terms of 3rd power....
 240 = 2^3*2*3*5 ....
this means k has to be div by each of above, base can be taken if anything to power of 3 is there as the other powers will get adjusted by the power of k, k^3..
since k is an integer, we cannot take k to be div by 3rd root of 3 and 5, it has to be div by 3 and 5..
so k^3 is div by  2^3*2*3*5. .
remove the 3rd powers..
so k will be div by 2*2*3*5 or 60..

hope it helped you with concept.

marzan2011 · Answer

hi Buneal your works are just grate

would you please explain 
 k^3=2^6*3^3*5^3
how power of 2 got 6

Posted from my mobile device

Bunuel · Answer

We have that:
(i)  k^3  is divisible by  2^4*3*5 ;
(ii) the powers of each prime of  k^3  must be a multiple of 3;

Now,  k^3  is divisible  2^4 , so the power of 2 in  k^3  must be multiple of 3 and more that 4: the least multiple of 3 which is more that 4 is 6.

Hope it's clear.

seabhi · Answer

Hi Bunuel,
To divide K^3. don't we need . 2^2*3^2*5^2.

Bunuel · Answer

Sorry don't follow you...

You mean that the least value of k must be 2^2*3^2*5^2? Why?

seabhi · Answer

Hi Bunuel,

Because to convert  240=2^4*3*5  to be divisible by  k^3  don't we need  2^2*3^2*5^2

Bunuel · Answer

We are told that k^3 is divisible by 240, not vise-versa. Also, what is 2^2*3^2*5^2? Is it the value of k, k^3, ...

ravishankar1788 · Answer

I just used 60 as an answer. Since this is a 600 level question, I guess using answer directly would make sense. Thoughts?

apple08 · Answer

Dear Bunuel 
How to get k=2^2*3*5=60  From  2^4*3*5  is  k^3=2^6*3^3*5^3  -->  k=2^2*3*5=60 .
Appreciate your reply 
Many thanks

DennisEisele · Answer

Dear Apple08

I assume you're doing the same mistake as seabhi. The question is asking for the  "least possible value of the integer  k " . It's not asking for  k^3 .

Of course, to end up with a perfect cube ( k^3 ) you need to add  2^2 * 3^2 * 5^2  but thats not questioned. Rephrasing the question then you're actually seeking for the 3rd root of ( k^3 ).
 \sqrt {2^6 * 3^3 * 5^3} = 2^2 * 3 * 5  and since  2^2 * 3 * 5  equals 60, the correct answer is C.

Hope it helped you
Greetings from Germany
Dennis

JeffTargetTestPrep · Answer

Since k^3/240 = integer, we can say that the product of 240 and some integer n is equal to a perfect cube. In other words, 240n = k^3.

We must remember that all perfect cubes break down to unique prime factors, each of which has an exponent that is a multiple of 3. So let’s break down 240 into primes to help determine what extra prime factors we need to make 240n a perfect cube.
 
240 = 24 x 10 = 8 x 3 x 2 x 5 = 2 x 2 x 2 x 3 x 2 x 5 = 2^4 x 3^1 x 5^1
 
In order to make 240n a perfect cube, we need two more 2s, two more 3s and two more 5s. Thus, the smallest perfect cube that is a multiple of 240 is 2^6 x 3^3 x 5^3.

To determine the least possible value of k, we can take the cube root of 2^6 x 3^3 x 5^3 and we have:

2^2 x 3 x 5 = 4 x 3 x 5 = 60

Thus, the minimum value of k is 60.

Answer: C

prathyushaR · Answer

given,
 k^3  is divisible by 240 which means  k^3  is a multiple of 240
i.e.  k^3 = 240 * (non zero integer 'n')  
 k^3= n* 24*10 =n*8*3*2*5 = n*2^4 *3*5  
 the least integer value of n will give least integer value of k, to make the above expression a perfect cube the least integer value of n is
 n=2^2 *3^2* 5^2, thus, k^3=2^6*3^3*5^3  
therefore least possible value of k is:
 2^2*3*5=60 
ANSWER: C

BrushMyQuant · Answer

↧↧↧ Weekly Video Solution to the Problem Series ↧↧↧

https://www.youtube.com/watch?v=N2VqhGJbJiE

Given that k^3 is divisible by 240 and we need to find the minimum value of k

\frac{k^3}{240} . Now we have  k^3  in the numerator so all terms in the numerator will have a power which is a multiple of 3

=> to find the least value of k we need to find the least value of  k^3  and to do that we need to factorize 240 and make all powers of the prime factors of 240 as a power of 3

240 =  2^4 * 3^1 * 5^1

For all powers to be multiples of 3 we can make them as  2^6 * 3^3 * 5^3

=>  k^3 = 2^6 * 3^3 * 5^3  =  (2^2 * 3 * 5)^3 
=> k =  2^2 * 3 * 5  = 4 * 15 = 60

So,  Answer will be C 
Hope it helps!

NEYR0N · Answer

240 = 2^4 * 3 *5

and k^3 divisible by 240 that means k^3 must either have all the primes of 240, or more.

its asking about K .. 12 = 2^2 * 3 ---- doesn't contain 5 , go to 30

30 = 3 * 2 * 5 , ok but k^3 yields 2^3 ... not enough we need one more 2 . so move to 60

60 = 2*3*2*5 . ok done. C

If k^3 is divisible by 240 what is the least possible value

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If k^3 is divisible by 240 what is the least possible value

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